Question

Determine the number of ways of placing the numbers $1, 2, 3, \dots, 9$ in a circle, so that the sum of any three numbers in consecutive positions is divisible by $3.$ (Two arrangements are considered the same if one arrangement can be rotated to obtain the other.)

Answer 1

Answer:

144 ways

Step-by-step explanation:

The arrangements to have the sum of three numbers being divisible by three can be approached by modular arithmetic such that the difference between the third integer and the sum of the first two integers is divisible by 3 making them congruent modulus 3

We have;

With the first number as 1

The second number can be 2, 5 or 8

The third number can then be 3, 6, 9

The fourth number can then be 4 or 7

The sixth number can be 3, 6, 9

The seventh number can be 4 or 7

The eight number can be 2, 5, 8

and the ninth number can be 3, 6, 9

Therefore, we have

the first number as 1

The second, fifth, and eight number can be 2, 5 or 8

The third, sixth and ninth number can then be 3, 6, 9

The fourth, and seventh number can then be 4 or 7 Which gives,

The number of possible arrangements for the numbers 2, 5, and 8 = 3! = 6

The number of possible arrangements for the numbers 3, 6, and 9 = 3! = 6

The number of possible arrangements for the numbers 4, and 7 = 2! = 2

he total number of possible arrangements = 2×6×6 =72

The arrangement can be reversed in the counter clockwise directions given  the total number of ways of placing the numbers 1, 2, 3,..., 9 in a circle, so that the sum of any three numbers in consecutive positions is divisible by 3 = 72 × 2 = 144 ways.

Answer 2

I hope that Brainly supports $\LaTeX$ code. If it doesn't, then I am sorry if you cannot read this:

First, we know that each number is 0, 1, or 2 modulo 3.

This table will help illustrate the idea:

$$\begin{array}{c|c}x&x\mod 3\\0&0\\1&1\\2&2\\3&0\\4&1\\5&2\\6&0\\7&1\\8&2\\9&0\end{array}$$

If the sum of 3 adjacent numbers is a multiple of 3, then the sum of the numbers modulo 3 must be 0.

This means that you can arrange the modulo of the numbers in the patterns shown below:

$0,1,2,0,1,2,0,1,2$

or

$2,1,0,2,1,0,2,1,0$

Note that these are the only two unique patterns, as patterns like $1,2,0,1,2,0,1,2,0$ is just $0,1,2,0,1,2,0,1,2$ rotated counterclockwise.

If we substitute each $0$ for a multiple of $3$ (3, 6, 9), each $1$ for $1$ more than a multiple of $3$ (1, 4, 7) and each $2$ for $2$ more than a multiple of $3$ (2, 5, 8), we will get a possible placement.

Now we can use a certain pattern, and we get that there are $3 \cdot 2 \cdot 1 \cdot 3 \cdot 2 \cdot 1 \cdot 3 \cdot 2 \cdot 1 = 6 \cdot 6 \cdot 6 = 216$ arrangements per pattern. Since there are $2$ patterns, we have $216 \cdot 2 = \boxed{432 \text{ total arrangements!}}$